Prove the equivalence of two definitions for an unramified homomorphism at a prime ideal by showing the relationship between localization, field extensions, and tensor product properties | Step-by-Step Solution

Hello! This is a beautiful problem in commutative algebra that connects several fundamental concepts.

1. What We're Solving:

We need to prove that different characterizations of "unramified homomorphisms" are actually equivalent. An unramified homomorphism is essentially one that behaves "nicely" - it doesn't create any "ramification" or "branching" when we extend from ring R to ring A.

2. The Approach:

The key insight is that "unramified" can be detected in multiple ways:

• Locally: by looking at what happens after localizing at prime p
• Residually: by examining the induced field extension
• Tangentially: by studying the cotangent space (via Ω₁)

We'll show these perspectives are equivalent by creating a logical chain of implications.

3. Step-by-Step Solution:

Step 1: Set up the main definitions For a homomorphism φ: R → A and prime p ∈ Spec(R), we have several candidate definitions for "unramified at p":

Definition 1 (Field extension condition): The induced map κ(p) → κ(q) is a finite separable field extension for all q ∈ Spec(A) lying over p.

Definition 2 (Localization condition): The localized map R_p → A_q is unramified for all such q.

Definition 3 (Cotangent condition): The A-module Ω₁_{A/R} ⊗_A κ(q) = 0 for all q over p.

Step 2: Show Definition 1 ⟹ Definition 2

• Start with the assumption that κ(p) → κ(q) is finite separable
• Use the fact that localization preserves the residue field structure
• The key insight: if the residue field extension is separable, then the local ring extension doesn't introduce new "infinitesimal directions"

Step 3: Show Definition 2 ⟹ Definition 3

• Use the fundamental exact sequence: I/I² → Ω₁_{R_p/R_p} ⊗ A_q → Ω₁_{A_q/R_p} → 0
• Here I is the kernel of the natural map R_p ⊗_R A_q → A_q
• When the map is unramified locally, this forces Ω₁_{A_q/R_p} = 0
• By base change properties, this gives us the vanishing of cotangent modules

Step 4: Show Definition 3 ⟹ Definition 1

• This is the most subtle direction!
• Use the fact that Ω₁_{A/R} measures "infinitesimal extensions"
• The vanishing of Ω₁_{A/R} ⊗_A κ(q) means there are no non-trivial infinitesimal deformations
• By the relationship between separability and differential forms, this forces the residue field extension to be separable
• The finiteness follows from our assumption that A is finitely generated over R

Step 5: Verify the equivalences are well-defined

• Check that our conditions are independent of the choice of q lying over p
• Use the fact that all primes lying over p will give equivalent conditions due to the Noetherian hypotheses

4. The Framework:

The complete equivalence gives us:

Theorem: For φ: R → A as above and p ∈ Spec(R), the following are equivalent:

• 1. φ is unramified at p (field extension definition)
• 2. φ is locally unramified at p (localization definition)
• 3. The cotangent module vanishes appropriately at p

This shows that unramifiedness is a robust concept that can be detected through algebra, geometry, or differential methods!

5. Memory Tip:

Think of "unramified" as meaning "no surprises":

• No surprise field extensions (they're separable)
• No surprise local behavior (localization behaves predictably)
• No surprise tangent directions (cotangent module vanishes)

The beauty is that checking any one of these conditions guarantees all the others! This is why unramified morphisms are so fundamental in algebraic geometry and number theory.

Would you like me to elaborate on any particular step or connection? The interplay between these different characterizations is really the heart of modern algebraic geometry!